Kitonum

13755 Reputation

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11 years, 150 days

MaplePrimes Activity


These are answers submitted by Kitonum

After with(geometry):  write assume(a,real): 

Maple does not know that you suspect this condition!

 

Code of procedure:

P := proc (Eq)

local z, u;

z:=rhs(Eq)^(1/op(2,lhs(Eq)));

u := evalc(z);

if is(0 <= Re(u)) and is(0 <= Im(u)) then abs(z)*(cos(argument(u))+I*sin(argument(u))) else abs(z)*(cos(argument(u)+2*Pi*op(2, z))+I*sin(argument(u)+2*Pi*op(2, z)));

end if;

end proc;

 

Example:

P(w^11=-5-5*I);

palindrome:=proc(n)

   option cache;

   uses ST = StringTools;

   if 

    modp(length(n),11)<>0 then

   return false;

   else

   return ST :- IsPalindrome(convert(n,'string'));

   end if;

end proc;

Use the seq command.

Example:

A:=seq(Vector([-2+3*i, -1+3*i, 3*i]), i=1..10):

ArrayTools:-Concatenate(2, A);

Here is a variant which suitable and for older versions of Maple (with only Classic Worksheet):

 

A:=plot([exp(-x),x^2], x=-0.5..1.5,color=black,thickness=2):

h:=fsolve(exp(-x)=x^2):

B:=plot(x^2,x=0..h,color=white,filled=true):

C:=plot(exp(-x),x=0..h,color=red,filled=true):

E:=plot([0, t, t=0..1], color=black, thickness=2):

plots[display](A,B,C,E,scaling=constrained,view=-0.3..1.7,labels=["x","y"]);

Yes, of course!

Example:

And what's the problem? Symbolically your system is solved correctly. This system can't be solved numerically, since it contains three parameters: v, va, f (t) .

Your function y(x)=-2*x^2+C1*ln(x)  is the solution of a differential equation

x*ln(x)*diff(y(x),x)-y(x)=2*x^2-4*x^2*ln(x)  which has no solutions for x<=0, since the domain of the function ln (x) only positive numbers. 

Write your last line as follows and get an answer:

Your decision will be correct if the line MN does not intersect the sphere S. Otherwise, the answer is 0. If the point M(1, 2, -1) then at the end of your code, add the command allvalues(%)

Let O be a center of the sphere.

1) Find the equation of the plane passing through the points M, N and O.

2) Find the equation of the line of intersection of this plane and the sphere.

3) The points for which the distance from the sphere to the line MN the maximum and minimum are on the line.

If n is known, then your equation is reduced to finding the roots of a polynomial of degree n. It is easy to prove that, for different  a_i, this polynomial will have n different real roots. But finding them in a closed form for large n seems impossible. See the example for n = 5:

restart:

f:=unapply(sum(1/(x+a[i]),i=1..n), (x,n));

n:=5;

a:=[seq(k,k=1..n)];

f(x,n);

solve(f(x,n)=1);

allvalues([%]);

evalf(%);

plot([f(x,n),1],x=-18..8,-3..5, color=[blue,red], thickness=2, discont=true);

Try this variant:

restart;

 k := 0; B[1] := 0.001/365; B[2] :=0.002/365; B[3] :=0.005/365:

ode := diff(U(t), t) = -(A[1]+A[2]*U(t))*U(t);

ic[0] := U(365*k) = 1000;

sol[0] := dsolve({ic[0], subs(A[1] = B[1], A[2] = B[2], ode)}, U(t), numeric);

sigma := .5;

for k to 4 do 

if k = 3 then A[1] := B[3]:

ode1 := diff(U(t), t) = -(A[1]+A[2]*U(t))*U(t):  end if;  tk := 365*k; 

V := rhs(sol[k-1](tk)[2]);  ic[k] := U(tk) = sigma*V;

if k=3 then sol[k] := dsolve({ic[k], subs(A[2] = B[2], ode1)}, U(t), numeric); else 

sol[k] := dsolve({ic[k], subs(A[1] = B[1], A[2] = B[2], ode)}, U(t), numeric); fi;

end do;

Maple already has such a procedure:

ceil(3.5);

     4

If i understand your problem, you're trying to experiment to find a straight line passing through the point (4, 4/3) and to be tangent to semicircle  y=-sqrt(1-x^2)+3 .  This problem can be solved exactly, as follows:  

1) At any point of the semicircle find the tangent line to this semicircle. You get the equation of a straight line, depending on the parameter x that specifies the position of the point of tangency.  

2) Using the condition that the line passes through the point  (4, 4/3) , find the point of tangency, and then your desired angle.

 

I got the angle phi=arctan((20-4*sqrt(10))/45)=0.1619229414

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